Solving Frouier Series Problem: Finding Partial Sums for Periodic Function

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[SOLVED] Frouier series problem

Lately I took Frouier series...

Homework Statement



Find the Fourier series of the function below which is assumed to have the period 2 [tex]\pi[/tex], and find the first three partial sums.

http://www.imagehosting.com/out.php/i1650739_untitled.JPG

Homework Equations

The Attempt at a Solution



I used office 2007 to write my solution...and here they are hosted as a picture..
http://www.imagehosting.com/out.php/i1650792_untitled1.JPG

I see my answer not correct, I hope someone helps me..Thanks
 
Last edited:
on Phys.org
You are correct to integrate from -pi to pi in the integral, but your function is non-zero over that entire interval, so I do not know why you have the function value as 0 in [tex][-\pi , -\pi/2][/tex] and [tex][\pi/2 , \pi][/tex].

Also I believe your constant in front of your integral for a_n is incorrect. The constant in front of that integral should be:

[tex]\frac{2}{Period}=\frac{2}{2\pi}=\frac{1}{\pi}[/tex]
 
Last edited:
sorry, there is a wrong in the picture of the function...the correction is
-2Pi becomes -Pi; -Pi becomes -Pi/2; Pi becomes P/2; and 2Pi becomes Pi.

so 'GO' the 0's in the integrals you specified becomes right with me..and you are correct about a_n, it should be 1/Pi and the same in the denominator of the cosine in that line..
 
hazim said:
sorry, there is a wrong in the picture of the function...the correction is
-2Pi becomes -Pi; -Pi becomes -Pi/2; Pi becomes P/2; and 2Pi becomes Pi.

so 'GO' the 0's in the integrals you specified becomes right with me..and you are correct about a_n, it should be 1/Pi and the same in the denominator of the cosine in that line..

OK your integration makes sense then.

BTW. I usually imagine my PF name being pronounces "G Zero One." :wink:

So, was the constant your problem or do are you still having trouble?
 
yes it was the constant..it's now solved..thanks
 
Glad to be of help.:smile:
 

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